Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.7 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.7

Question 1.

Solve the following Linear differential equations.

cos x \(\frac { dy }{ dx }\) + y sin x = 1

Solution:

The given differential equation can be written as

This is the form \(\frac { dy }{ dx }\) + py = Q

where P = tan x

Q = sec x

Thus, the given differential equation is linear.

I.F = e^{∫ pdx} = e^{∫ tan x dx} = e^{log (sec x)} = sec x

So, the required solution is given by

[y × I.F] = ∫ [Q × IF] dx + c

y × sec x = ∫ sec x × sec x dx + c

y sec x = ∫ sec² x dx + c

y sec x = tan x + c

= sin x + c cos x

y = sin x + c cos x is the required solution.

Question 2.

(1 – x²)\(\frac { dy }{ dx }\) – xy = 1

Solution:

The given differential equation can be written as

Thus, the given differential equation is linear.

Which is a required solution.

Question 3.

\(\frac { dy }{ dx }\) + \(\frac { y }{ x }\) = sin x

Solution:

The given differential equation can be written as

\(\frac { dy }{ dx }\) + (\(\frac { 1 }{ x }\))y = sin x

This is of the form \(\frac { dy }{ dx }\) + Py = Q

where P = \(\frac { 1 }{ x }\)

Q = sin x

Thus, the given differential equation is linear.

I.F= e^{∫ pdx} = e^{∫ \(\frac { 1 }{ x }\) dx} = e^{log x} = x

So, the required solution is given by

yx I.F = ∫ (Q × I.F) dx + c

yx = ∫ sin x × x dx + c

= x (-cos x) – (1) (-sin x) + c

yx = -x cos x + sin x + c

yx + x cos x = sin x + c

(y + cos x) x = sin x + c is a required solution.

Question 4.

(x² + 1)\(\frac { dy }{ dx }\) + 2xy = \(\sqrt { x^2+4 }\)

Solution:

The given differential equation may be written as

Thus, the given differential equation is linear.

I.F= e^{∫ pdx} = e^{∫ \(\frac { 2x }{ x^+1 }\) dx} = e^{log (x² + 2)} = x² + 1

So, the required solution is given by

Is the required solution.

Question 5.

(2x – 10y³) dy + y dx = 0

Solution:

The given differential equation may be written as

y dx = -(2x – 10y³) dy

So, the required solution is

x × I.F = ∫ (Q × I.F)dy + c

xy² = ∫ 10 y² × y² dy + c

= ∫ 10 y^{4} dy + c

= \(\frac { 10y^5 }{ 5 }\) + c = 2y^{5} + c

xy² = 2y^{5} + c is a required solution.

Question 6.

x sin x \(\frac { dy }{ dx }\) + (x cos x + sin x) y = sin x

Solution:

The given differential equation can be written as

Thus, the given differential equation is linear.

I.F = e^{∫ pdx} = e^{∫ (cot x + 1/x)dx} = e^{log sin x + log x}

= e^{log (x sin x)} = x sin x

So, the solution of the given differential equation is given by

y × I.F = ∫(Q × I.F) dx + c

y (x sin x) = ∫ \(\frac { 1 }{ x }\) x sin x dx + c

= ∫ sin x dx + c

y (x sin x) = -cos x + c

xy sin x + cos x = c is the required solution.

Question 7.

Solve (y – e^{sin-1x}) \(\frac { dx }{ dy }\) + \(\sqrt { 1-x^2 }\) = 0

Solution:

Thus, the given differential equation is Linear

Question 8.

\(\frac { dy }{ dx }\) + \(\frac { y }{ (1-x)√x }\) = 1 – √x

Solution:

The given linear differential equation is of the form

\(\frac { dy }{ dx }\) + py = Q

Question 9.

(1 + x + xy²) \(\frac { dy }{ dx }\) + (y + y³) = 0

Solution:

So, the solution of the equation is given by

xy + tan^{-1} y = c

Which is the required solution.

Question 10.

\(\frac { dy }{ dx }\) + \(\frac { y }{ x log x }\) = \(\frac { sin 2x }{ log x }\)

Solution:

The given differential equation may be written as

\(\frac { dy }{ dx }\) + \(\frac { 1 }{ x log x }\)y = \(\frac { sin 2x }{ log x }\)

This is of the form \(\frac { dy }{ dx }\) + Py = Q

Where P = \(\frac { 1 }{ x log x }\); Q = \(\frac { sin 2x }{ log x }\)

Thus, the differential equation is linear.

I.F = e^{∫pdx}

= e^{∫ \(\frac { 1 }{ x log x }\) dx}

= e^{∫ \(\frac { 1 }{ t }\) dt}

= e^{log t}

= log x

So, the solution of the given differential equation is

y × I.F = ∫ (Q × I.F) dx + c

y × log x = ∫ \(\frac { sin 2x }{ log x }\) × log x dx + c

= ∫ sin 2x dx + c

y log x = \(\frac { -cos 2x }{ 2 }\) + c

y log x + \(\frac { cos 2x }{ 2 }\) = c is a required solution.

Question 11.

(x + a) \(\frac { dy }{ dx }\) – 2y = (x + a)^{4}

Solution:

= -2 log (x + a)

= log (x + a)^{-2}

Question 12.

\(\frac { dy }{ dx }\) = \(\frac { sin^x }{ 1+x^3 }\) – \(\frac { 3x^2 }{ 1+x^3 }\)y

Solution:

The equation can be written as

Question 13.

x \(\frac { dy }{ dx }\) + y = x log x

Solution:

The given differential equation may be written as

Thus, the given differential equation is linear.

I.F = e^{∫pdx} = e^{∫\(\frac { 1 }{ x }\) dx} = e^{log x} = x

So, the solution of the given differential equation is given by

y × I.F = ∫(Q × I.F)dx + c

yx = ∫ log x x dx + c

yx = ∫ x log x dx + c

u = log x ∫dv = ∫x dx

Multiply by 4

4yx = 2x² log x – x² + 4c

4xy = 2x² log x – x² + 4c is a required solution.

Question 14.

x \(\frac { dy }{ dx }\) + 2y – x² log x = 0

Solution:

The given differential equation may be written as

\(\frac { x }{ x }\) \(\frac { dy }{ dx }\) + \(\frac { 2y }{ x }\) = \(\frac { x log x }{ x }\)

This is of the form \(\frac { dy }{ dx }\) + Py = Q

where P = \(\frac { 2 }{ x }\); Q = x log x

Thus, the given equation is linear.

I.F = e^{∫pdx} = e^{∫\(\frac { 2 }{ x }\) dx}

e^{log x} = e^{log x²} = x²

So the required solution is

y × I.F = ∫(Q × I.F) dx + c

yx² = ∫x log x x² dx + c

yx² = ∫x³ logx dx + c

Multiply by 16

16x²y = 4x^{4} log x – x^{4} + 16c is a required solution

Question 15.

\(\frac { dy }{ dx }\) + \(\frac { 3y }{ x }\) = \(\frac { 1 }{ x^2 }\), given that y = 2 when x = 1

Solution:

The given differential equation can be written as

So, its solution is given by

y × I.F = ∫(Q × I.F) dx + c

yx³ = ∫ \(\frac { 1 }{ x^2 }\) + x³ dx + c

= ∫x dx + c

y x³ = \(\frac { x^2 }{ 2 }\) + c

2yx³ = x² + c

Given that y = 2 when x = 1

2 (2) (1)³ = 1 + c

4 – 1 = c

c = 3

∴ 2yx³ = x² + 3 is a required solution.